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Probability
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4.3
Rules of Probability
Copyright © Cengage Learning. All rights reserved.
Finding the Probability of “Not A”
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Finding the Probability of “Not A”
The concept of complementary events is fundamental to
finding the probability of “not A.”
In complementary events, the complement of an event A,
, is the set of all sample points in the sample space that
do not belong to event A.
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Finding the Probability of “Not A”
A few examples of complementary events are
(1) the complement of the event “success” is “failure,”
(2) the complement of “selected voter is Republican” is
“selected voter is not Republican,” and
(3) the complement of “no heads” on 10 tosses of a coin is
“at least one head.”
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Finding the Probability of “Not A”
By combining the information in the definition of
complement with Property 2 (“The sum of the probabilities
for all outcomes of an experiment is equal to exactly one.”),
we can say that
As a result of this relationship, we have the complement
rule, which states that the probability of A complement
= one minus the probability of A.
This rule can be expressed algebraically this way:
(4.3)
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Finding the Probability of “Not A”
Every event A has a complementary event
.
Complementary probabilities are very useful when the
question asks for the probability of “at least one.”
Generally, this represents a combination of several events,
but the complementary event “none” is a single outcome.
It is easier to solve for the complementary event and get
the answer by using formula (4.3).
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Finding the Probability of “Not A”
Using Complements to Find Probabilities
One way to find probabilities is to use complements. Two
dice are rolled.
What is the probability that the sum is at least 3 (that is, 3,
4, 5, ... , 12)?
Suppose one of the dice is black and the other is white.
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Finding the Probability of “Not A”
Rather than finding the probability for each of the sums 3,
4, 5, ... , 12 separately and adding, it is much simpler to
find the probability that the sum is 2 (“less than 3”) and then
use formula (4.3) to find the probability of “at least 3,”
because “less than 3” and “at least 3” are complementary
events.
Using formula (4.3),
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Finding the Probability of “A OR B”
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Finding the Probability of “A OR B”
An hourly wage earner wants to estimate the chances of
“receiving a promotion or getting a pay raise.” The worker
would be happy with either outcome.
Historical information is available that will allow the worker
to estimate the probability of “receiving a promotion” and
“getting a pay raise” separately.
In this section we will learn how to apply the addition rule
to find the compound probability of interest.
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Finding the Probability of “A OR B”
General Addition Rule
Let A and B be two events defined in a sample space, S.
In words:
probability of A or B = probability of A + probability of B
– probability of A and B
In algebra:
P(A or B) = P(A) + P(B) – P(A and B)
(4.4)
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Finding the Probability of “A OR B”
To see if the relationship expressed by the general addition
rule works, let’s look at it in action.
A statewide poll of 800 registered voters in 25 precincts
from across New York state was taken.
Each voter was identified as being registered as
Republican, Democrat, or other and then asked, “Are you
in favor of or against the current budget proposal awaiting
the governor’s signature?”
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Finding the Probability of “A OR B”
The resulting tallies are shown here.
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Finding the Probability of “A OR B”
Suppose one voter is to be selected at random from the
800 voters summarized in the preceding table.
Let’s consider the two events “The voter selected is in
favor” and “The voter is a Republican.”
Find the four probabilities: P(in favor), P(Republican),
P(in favor or Republican), and P(in favor and Republican).
Then use the results to check the truth of the addition rule.
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Finding the Probability of “A OR B”
We find that the:
• Probability the voter selected is “in favor”
= P(in favor) = 464/800 = 0.58.
• Probability the voter selected is “Republican”
= P(Republican) = 224/800 = 0.28.
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Finding the Probability of “A OR B”
• Probability the voter selected is “in favor” or “Republican”
= P(in favor or Republican)
= (136 + 314 + 14 + 88)/800 = 552/800 = 0.69.
• Probability the voter selected is “in favor” and
“Republican”
= P(in favor and Republican)
= 136/800 = 0.17.
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Finding the Probability of “A OR B”
Notes about Finding the Preceding Probabilities:
1. The connective “or” means “one or the other or both”;
thus, “in favor or Republican” means all voters who
satisfy either event.
2. The connective “and” means “both” or “in common”;
thus, “in favor and Republican” means all voters who
satisfy both events.
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Finding the Probability of “A OR B”
Now let’s use the preceding probabilities to demonstrate
the truth of the addition rule.
Let A = “in favor” and B = “Republican.” The general
addition rule then becomes:
P(in favor or Republican) = P(in favor) + P(Republican)
– P(in favor and Republican)
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Finding the Probability of “A OR B”
Remember, previously we found P(in favor or Republican)
= 0.69.
Using the other three probabilities, we see:
P (in favor) + P(Republican) – P(in favor and Republican)
= 0.58 + 0.28 – 0.17
= 0.69
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Finding the Probability of “A OR B”
Thus, we obtain identical answers by applying the addition
rule and by referring to the relevant cells in the table.
You typically do not have the option of finding P(A or B) two
ways, as we did here. You will be asked to find
P(A or B) starting with P(A) and P(B).
However, you will need a third piece of information. In the
previous situation, we needed P(A and B). We will need to
know either P(A and B) or some information that allows us
to find it.
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Finding the Probability of
“A AND B”
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Finding the Probability of “A AND B”
Suppose a criminal justice professor wants his class to
determine the likeliness of the event “a driver is ticketed for
a speeding violation and the driver had previously attended
a defensive driving class.”
The students are confident they can find the probabilities of
“a driver being ticketed for speeding” and “a driver who has
attended a defensive driving class” separately.
In this section we will learn how to apply the multiplication
rule to find the compound probability of interest.
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Finding the Probability of “A AND B”
General Multiplication Rule
Let A and B be two events defined in sample space S.
In words:
probability of A and B = probability of A  probability
of B, knowing A
In algebra:
P(A and B) = P(A)  P(B | A)
(4.5)
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Finding the Probability of “A AND B”
Note
When two events are involved, either event can be
identified as A, with the other identified as B.
The general multiplication rule could also be written as
P(B and A) = P(B)  P(A | B).
To have a better understanding of the multiplication rule,
let’s go back to our statewide poll of 800 registered voters
in 25 precincts from across New York state.
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Finding the Probability of “A AND B”
Each voter was identified as being registered as
Republican, Democrat, or other and then asked, “Are you
are in favor of or against the current budget proposal
awaiting the governor’s signature?”
The resulting tallies are shown here.
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Finding the Probability of “A AND B”
Suppose one voter is to be selected at random from the
800 voters summarized in the preceding table.
Let’s consider the two events: “The voter selected is in
favor” and “The voter is a Republican.”
Find the three probabilities: P(in favor), P(Republican | in
favor), and P(in favor and Republican).
Then use the results to check the truth of the multiplication
rule.
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Finding the Probability of “A AND B”
We find that the:
• Probability the voter selected is “in favor”
= P(in favor) = 464/800 = 0.58.
• Probability the voter selected is “Republican, given in
favor”
= P(Republican | in favor)
= 136/464 = 0.29.
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Finding the Probability of “A AND B”
• Probability the voter selected is “in favor” and
“Republican”
= P(in favor and Republican)
= 136/800 =
= 0.17.
Notes About Finding the Preceding Probabilities
1. The conditional “given” means there is a restriction; thus,
“Republican | in favor” means we start with only those
voters who are “in favor.” In this case, this means we are
looking only at 464 voters when determining this
probability.
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Finding the Probability of “A AND B”
2. The connective “and” means “both” or “in common”;
thus, “in favor and Republican” means all voters who
satisfy both events.
Now let’s use the previous probabilities to demonstrate the
truth of the multiplication rule.
Let A = “in favor” and B = “Republican.” The general
multiplication rule then becomes:
P(in favor and Republican)
= P(in favor)  P(Republican | in favor)
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Finding the Probability of “A AND B”
Previously we found:
P(in favor and Republican) =
= 0.17.
Using the other two probabilities, we see:
P(in favor)  P(Republican | in favor)
= 0.17.
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Finding the Probability of “A AND B”
Conditional Probability And Drawing Without
Replacement
You typically do not have the option of finding P(A and B)
two ways, as we did here. When you are asked to find P(A
and B), you will often be given P(A) and P(B).
However, you will not always get the correct answer by just
multiplying those two probabilities together.
You will need a third piece of information: the conditional
probability of one of the two events or information that will
allow you to find it.
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Finding the Probability of “A AND B”
At a carnival game, the player blindly draws one colored
marble at a time from a box containing two red and four
blue marbles.
The chosen marble is not returned to the box after being
selected; that is, each drawing is done without
replacement. The marbles are mixed before each drawing.
It costs $1 to play, and if the first two marbles drawn are
red, the player receives a $2 prize.
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Finding the Probability of “A AND B”
If the first four marbles drawn are all blue, the player
receives a $5 prize. Otherwise, no prize is awarded.
To find the probability of winning a prize, let’s look first at
the probability of drawing red or blue on consecutive
drawings and organize the information on a tree diagram.
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Finding the Probability of “A AND B”
On the first draw (represented by the purple branch
segments in Figure 4.4), the probability of red is two
chances out of six, , or , whereas the probability
of blue is , or .
Tree Diagram—First Two Drawings, Carnival Game
Figure 4.4
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Finding the Probability of “A AND B”
Because the marble is not replaced, only five marbles are
left in the box; the number of each color remaining depends
on the color of the first marble drawn.
If the first marble was red, then the probabilities are and
, as shown on the tree diagram (green branch segments
in Figure 4.4).
If the first marble was blue, then the probabilities are and
, as shown on the tree diagram (orange branch segments
in Figure 4.4).
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Finding the Probability of “A AND B”
The probabilities change with each drawing, because the
number of marbles available keeps decreasing as each
drawing takes place.
The tree diagram is a marvelous pictorial aid in following
the progression.
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Finding the Probability of “A AND B”
The probability of winning the $2 prize can now be found
using formula (4.5):
P(A and B) = P(A)  P(B | A)
P(winning $2) = P(R1 and R2)
= P(R1)  P(R2 | R1)
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Finding the Probability of “A AND B”
Note
The tree diagram, when labeled, has the probabilities
needed for multiplying listed along the branch representing
the winning effort.
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